(a) Consider the population 3, 4, 9, 14, and 20. The *z*-scores for each of the five data values are 

*z*3 ≈ −1.101,
 
*z*4 ≈ −0.944,
 
*z*9 ≈ −0.157,
 
*z*14 ≈ 0.629,
 
and *z*20 ≈ 1.573.

 Find the mean and the population standard deviation of 

*z*3,
 
*z*4,
 
*z*9,
 
*z*14,
 
and *z*20.

 (Round answers to the nearest whole number.)


mean    

standard deviation    


(b) Consider the population 2, 6, 12, 17, 22, and 25. The *z*-scores for each of the six data values are 

*z*2 ≈ −1.459,
 
*z*6 ≈ −0.973,
 
*z*12 ≈ −0.243,
 
*z*17 ≈ 0.365,
 
*z*22 ≈ 0.973,
 
and *z*25 ≈ 1.337.

 Find the mean and the population standard deviation of 

*z*2,
 
*z*6,
 
*z*12,
 
*z*17,
 
*z*22,
 
and *z*25.

 (Round answers to the nearest whole number.)


mean    

standard deviation    


### Part (a)

To find the mean and standard deviation of the given z-scores:

\[ z3 \approx -1.101 \]
\[ z4 \approx -0.944 \]
\[ z9 \approx -0.157 \]
\[ z14 \approx 0.629 \]
\[ z20 \approx 1.573 \]

#### Mean of z-scores

The mean (\(\mu_z\)) of the z-scores is given by the average of the z-scores:

\[ \mu_z = \frac{z3 + z4 + z9 + z14 + z20}{5} \]

Substitute the given values:

\[ \mu_z = \frac{-1.101 - 0.944 - 0.157 + 0.629 + 1.573}{5} \]

\[ \mu_z = \frac{-1.101 - 0.944 - 0.157 + 0.629 + 1.573}{5} \]

\[ \mu_z = \frac{-1.573 + 2.202}{5} \]

\[ \mu_z = \frac{0.629}{5} \]

\[ \mu_z = 0.1258 \]

\[ \mu_z \approx 0 \, (\text{rounded to the nearest whole number}) \]

#### Standard Deviation of z-scores

The standard deviation (\(\sigma_z\)) of the z-scores is given by:

\[ \sigma_z = \sqrt{\frac{(z3 - \mu_z)^2 + (z4 - \mu_z)^2 + (z9 - \mu_z)^2 + (z14 - \mu_z)^2 + (z20 - \mu_z)^2}{5}} \]

Since \(\mu_z \approx 0\):

\[ \sigma_z = \sqrt{\frac{(-1.101)^2 + (-0.944)^2 + (-0.157)^2 + (0.629)^2 + (1.573)^2}{5}} \]

\[ \sigma_z = \sqrt{\frac{1.212 + 0.891 + 0.025 + 0.396 + 2.474}{5}} \]

\[ \sigma_z = \sqrt{\frac{4.998}{5}} \]

\[ \sigma_z = \sqrt{0.9996} \]

\[ \sigma_z \approx 1 \, (\text{rounded to the nearest whole number}) \]

### Part (b)

To find the mean and standard deviation of the given z-scores:

\[ z2 \approx -1.459 \]
\[ z6 \approx -0.973 \]
\[ z12 \approx -0.243 \]
\[ z17 \approx 0.365 \]
\[ z22 \approx 0.973 \]
\[ z25 \approx 1.337 \]

#### Mean of z-scores

The mean (\(\mu_z\)) of the z-scores is given by the average of the z-scores:

\[ \mu_z = \frac{z2 + z6 + z12 + z17 + z22 + z25}{6} \]

Substitute the given values:

\[ \mu_z = \frac{-1.459 - 0.973 - 0.243 + 0.365 + 0.973 + 1.337}{6} \]

\[ \mu_z = \frac{-2.675 + 2.675}{6} \]

\[ \mu_z = \frac{0}{6} \]

\[ \mu_z = 0 \]

#### Standard Deviation of z-scores

The standard deviation (\(\sigma_z\)) of the z-scores is given by:

\[ \sigma_z = \sqrt{\frac{(z2 - \mu_z)^2 + (z6 - \mu_z)^2 + (z12 - \mu_z)^2 + (z17 - \mu_z)^2 + (z22 - \mu_z)^2 + (z25 - \mu_z)^2}{6}} \]

Since \(\mu_z \approx 0\):

\[ \sigma_z = \sqrt{\frac{(-1.459)^2 + (-0.973)^2 + (-0.243)^2 + (0.365)^2 + (0.973)^2 + (1.337)^2}{6}} \]

\[ \sigma_z = \sqrt{\frac{2.129 + 0.947 + 0.059 + 0.133 + 0.947 + 1.788}{6}} \]

\[ \sigma_z = \sqrt{\frac{6.003}{6}} \]

\[ \sigma_z = \sqrt{1.0005} \]

\[ \sigma_z \approx 1 \, (\text{rounded to the nearest whole number}) \]

#### Summary

**Part (a):**
- Mean: 0
- Standard Deviation: 1

**Part (b):**
- Mean: 0
- Standard Deviation: 1

Do you want details or have any questions? 

Here are 5 numbered questions you can ask:

1. How do you interpret a z-score in statistics?
2. Can you explain how to calculate the mean and standard deviation?
3. What is the significance of the standard deviation in a data set?
4. How do you find the probability corresponding to a z-score?
5. What are the properties of the normal distribution?

**Tip:** The mean of a set of z-scores for a standard normal distribution is always 0, and the standard deviation is always 1.

Learn how to calculate the mean and standard deviation of z-scores for given data sets. Detailed step-by-step solution with rounded answers to the nearest whole number. Suitable for undergraduate students studying statistics.

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